Convolution of independent random variables

If independent real random variables $X$ and $Y$ have densities $f_X$ and $f_Y$, their sum has the convolution density

Let $X$ take integer values with probabilities $p_n$, and let $Y_{\sigma }$ be an independent centered normal distribution with positive standard deviation $\sigma$. Conditioning on $X$ gives the probability density function of their sum as $g_{\sigma }(z)={\sum }_n{p}_n{\phi }_{\sigma }(z-n)$, where ${\phi }_{\sigma }$ is the normal density. Nonnegativity and Tonelli's theorem show that this density integrates to one. For small $\sigma$, isolated atoms become narrow smooth peaks with their original probability masses.

For each fixed real $t$, dominated convergence givessince $X/\sigma \to 0$ almost surely and ${\phi }_1$ is bounded. Thus every fixed integer-valued distribution acquires a Gaussian central profile on this expanding scale, without any moment assumption on $X$. This is not a claim about uniform relative error in the tails or about finite moments. For example, if $p_n=3/({\pi }^2{n}^2)$ for $n\ne 0$ and $p_0=0$, the expectation of $|X|$ is infinite. Since $\mathbb{E}|Y_{\sigma }|<\infty$, the triangle inequality forces $\mathbb{E}|X+Y_{\sigma }|=\infty$ too. Symmetry then gives equal infinite positive and negative first moments, so there is no ordinary mean. The second moment is also infinite. A Gaussian-looking center can therefore coexist with a heavy-tailed distribution.

Let $X$ have the Standard Cauchy distribution and let $Y$, independently, have the uniform distribution on $[-h,h]$, where $h>0$. The convolution of probability densities gives the probability density of $Z=X+Y$ asThis is an even function. Its derivative isso its unique mode is zero. Its tails satisfy $p_h(z)\sim 1/(\pi z^2)$ as $|z|\to \infty$: multiplying the integral by $z^2$ makes the integrand converge uniformly to $1$ on the bounded interval $[-h,h]$. Consequently both positive and negative parts of the first moment diverge, so its expected value does not exist. The symmetric Cauchy principal value of that moment is zero, which is a different assertion. Adding a bounded independent displacement smooths the probability density without removing the divergent first moment of the Cauchy distribution.

For a taxi starting at any vertex of a rectangle with side lengths $s,t$, a uniform pickup vertex gives approach-distance law uniform on $0,s,t,s+t$. Conditional on that pickup, a uniform distinct destination gives passenger-distance law uniform on $s,t,s+t$, independent of the pickup vertex. Hence the two distance variables are independent despite sharing that vertex. Their convolution has twelve equally likely pairs. Its mean is $7(s+t)/6$ and its variance is $(s^2+t^2)/4+2(s^2+t^2-st)/9$. Repeated numerical sums combine their multiplicities when some distances coincide.